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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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then
Y'(t) = eat [u'(t) + au (t)]
and
Y"(t) = eat [u"(t) + 2au'(t) + a2u (t)].
Substituting for y, y, and Ó in Eq. (27), canceling the factor eat, and collecting terms, we obtain
au"(t) + (2aa + b)u'(t) + (aa2 + ba + c)u (t) = Pn (t). (28)
The determination of a particular solution ofEq. (28) is precisely the same problem, except for the names of the constants, as solving Eq. (24). Therefore, if aa2 + ba + c is not zero, we assume that u(t) = A0tn + ... + An; hence a particular solution of Eq. (27) is of the form
Y (t) = eat (A0tn + A1tn-1 + ... + An). (29)
On the other hand, if aa2 + ba + c is zero, but 2aa + b is not, we must take u(t) to be of the form t(A0tn + ... + An). The corresponding form for Y(t) is t times the expression on the right side ofEq. (29). Note that if aa2 + ba + c is zero, then eat is a solution of the homogeneous equation. If both aa2 + ba + c and 2aa + b are zero (and this implies that both eat and teat are solutions of the homogeneous equation), then the correct form for u(t) is t2( A0tn + ... + An). Hence Y(t) is t2 times the expression on the right side ofEq. (29).
g(t) = eatPn(t) cos â t or eatPn(t) sin â t. These two cases are similar, so we consider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Euler’s formula, sin â t = (e‘et - e~‘et )/2i. Hence g(t) is ofthe form
e(a+ie)t e(a-ie)t
g(t) = Pn (t)-----------2i--------
and we should choose
Y(t) = e(a+ie)t (A0tn + ... + An) + e(a-ie)t (B0tn + ... + Bn), or equivalently,
Y(t) = eat(A0tn + ... + An) cos ât + eat(B0tn + ... + Bn) sin â t.
Usually, the latter form is preferred. If a ± ³â satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one.
If the nonhomogeneous function involves both cos â t and sin â t, it is usually convenient to treat these terms together, since each one individually may give rise to the
178
Chapter 3. Second Order Linear Equations
same form for a particular solution. For example, if g(t) = t sin t + 2cos t, the form for Y(t) would be
Y(t) = (A0t + A.) sin t + (B0t + Bx) cos t, provided that sin t and cos t are not solutions of the homogeneous equation.
PROBLEMS
In each of Problems 1 through 12 find the general solution of the given differential equation.
1. /- 2/- 3y = 3e^
3. /- 2/- 3y = —3te-t
5. / + 9y = t2e3t + 6 7. 2/' + 3/ + y = t2 + 3 sin t
9. u" + o^u = cos ot, î2 = «2
11.
12.
Ó' + y + 4 y = 2 sinh t y" - Ó - 2y = cosh2t
2. Ó + 2 Ó + 5 y = 3 sin 2t
4. Ó' + 2 Ó = 3 + 4sin2t
6. Ó' + 2 y + y = 2e-t
8. Ó' + y = 3 sin2t + t cos2t
10. è11 + o'2 u = cos o0t
(e‘ - å- ' )/2
(ª + å -t )/2
In each of Problems 13 through 18 find the solution of the given initial value problem.
13. / + /- 2y = 2t, y(0) = 0, Ó (0) = 1
14. y" + 4y = t2 + 3e', y(0) = 0, Ó(0) = 2
15. /- 2/ + y = tet + 4, y(0) = 1, Ó (0) = 1
16. Ó- 2/- 3 y = 3te2t, y(0) = 1, Ó(0) = 0
17. Ó + 4y = 3 sin2t, y(0) = 2, /(0) = -1
18. /' + 2/ + 5y = 4e-tcos2t, y(0) = 1, /(0) = 0
In each of Problems 19 through 26:
(a) Determine a suitable form for Y(t) if the method of undetermined coefficients is to be used.
(b) Use a computer algebra system to find a particular solution of the given equation.
>
>
>
>
>
>
>
19.
21.
22. y
23. /
24. /
25. y
26. Ó
+ 3y = 2t4 + t2e-3t + sin3t > 20. Ó + y = t(1 + sint)
- 5y + 6y = e‘ cos2t + e2(3t + 4) sin t
+ 2y + 2y = 3e-t + 2e-t cos t + 4e-t^ sin t
- 4Ó + 4y = 2^ + 4te2t + tsin2t + 4y = t2 sin2t + (6t + 7) cos2t
+ 3y + 2y = º'(^ + 1) sin2t + 3e-t cos t + 4e‘
+ 2Ó + 5y = 3te-t cos2t - 2te-2t cos t
27. Determine the general solution of
Ó + ë2 y = J]
sin mn t,
m=1
where ê > 0 and Ë = mn for m = 1N.
> 28. In many physical problems the nonhomogeneous term may be specified by different formulas in different time periods. As an example, determine the solution y = ô(³) of
/ + y =
t,
ï ªÏ-t,
0 < t < ï, t > ï,
satisfying the initial conditions y(0) = 0 and Ó (0) = 1. Assume that y and y are also continuous at t = ï. Plot the nonhomogeneous term and the solution as functions of time. Hint: First solve the initial value problem for t < ï; then solve for t > ï, determining the constants in the latter solution from the continuity conditions at t = ï.
3.7 Variation of Parameters
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